Problem: Let $g(x)=\cos(\ln(3x))$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\sin(\ln(3x))}{x}$ (Choice B) B $3\sin(\ln(3x))$ (Choice C) C $\dfrac{\sin(\ln(3x))}{3x}$ (Choice D) D $-\sin(\ln(3x))$
$g$ is a composition of three functions! Let... $u(x)=\cos(x)$ $v(x)=\ln(x)$ $w(x)=3x$... then $g(x)=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $g'(x)$, we will need to use the chain rule twice! $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=-\sin(x)$ $v'(x)=\dfrac{1}{x}$ $w'(x)=3$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={-\sin(\ln(3x))}\cdot{ \dfrac{1}{3x}}\cdot{3} \\\\ &=-\dfrac{\cancel{3}\sin(\ln(3x))}{\cancel{3}x} \\\\ &=-\dfrac{\sin(\ln(3x))}{x} \end{aligned}$ In conclusion, $g'(x)=-\dfrac{\sin(\ln(3x))}{x}$.